Find $\lim_{x\to 0}(x+\sin x)^{\tan x}$ Find the limit $\lim_{x\to 0}(x+\sin x)^{\tan x}.$
I tried:
$\lim_{x\to 0}{\tan x}.\ln(x+\sin x)=\lim_{x\to 0}\frac{\ln(x+\sin x)}{\cot x}=\lim_{x\to 0}\frac{(1+\cos x)(\cos^2 x)}{-(x+\sin x)}$ 
I stuck at this step.
 A: Hints:
$$\left(x+\sin x)\right)^{\tan x}=e^{\tan x\log(x+\sin x)}$$
And now, using l'Hospital's rule"
$$\lim_{x\to 0}\frac{\log(x+\sin x)}{\cot x}\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{\frac{1+\cos x}{x+\sin x}}{-\frac1{\sin^2x}}$$
And 
$$\lim_{x\to 0}\frac{\sin^2x}{x+\sin x}\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{2\sin x\cos x}{1+\cos x}=0$$
Deduce now that the limit is $\;1\;$ using continuity of the exponential function
A: You did a mistake in the last identity, it should be $$\lim_{x\to 0}\frac{\ln(x+\sin x)}{\cot x}=\lim_{x\to 0}\frac{(1+\cos x)(-\sin^2 x)}{-(x+\sin x)}$$
And after that another L'Hopital gives $$\lim_{x\to 0}\frac{(1+\cos x)(-\sin^2 x)}{-(x+\sin x)}=\lim_{x\to 0}\frac{-\sin^3x + (1+\cos x)2\sin x\cos x}{1+\cos x}=0$$
A: $$\lim_{x\to0}{\tan x\ln(x+\sin x)}={\lim_{x\to0}\dfrac{\ln(x+\sin x)}{\cot x}}\stackrel{\text{L'H}}={\lim_{x\to0}\dfrac{\dfrac{1+\cos x}{x+\sin x}}{-\dfrac1{\sin^2x}}}=\lim_{x\to0} \left( -\dfrac{\sin^2x(1+\cos x)}{x+\sin x} \right) \stackrel{\text{L'H}}=\lim_{x\to0}{\dfrac{\sin^3x-2\sin x\cos x(\cos x+1)}{\cos x+1}}={\lim_{x\to0}\sin x(1-3\cos x)}=0$$
A: It's simpler with equivalents:
$$\tan x \sim_0 x,\quad\ln(x+\sin x)=\ln x + \ln\Bigl(1+\dfrac{\sin x}x\Bigr)\sim_0 \ln x $$
hence $\,\,\tan x\ln(x+\sin x)\sim_0 x\ln x$ so that $\,\,\tan x\ln(x+\sin x)\underset{x\to 0}{\longrightarrow}0\,\, $ and finally $$(x+\sin x)^{\tan x}\underset{x\to 0}{\longrightarrow} 1.$$
A: $$(x+\sin(x))^{\tan(x)}=e^{\tan(x)\ln(x+\sin(x))}$$
Compute $$\lim_{x\to0}\frac{\ln(x+\sin(x))}{\frac{1}{\sin(x)}}=\lim_{x\to0}\frac{\frac{1+\cos(x)}{x+\sin(x)}}{\frac{-\cos(x)}{\sin^2(x)}}=-2\lim_{x\to0}\frac{\sin(x)}{\frac{x}{\sin(x)}+1}=0$$
And use above.
A: This is merely a warning and a tip of intuition than an answer. It became too long for just being a comment. If this is considered bad habit, please tell me, and I'll remove it.
Firstly a comment on the existing solutions. I think that one should be careful with $\ln(x+\sin x)$, since $x+\sin x<0$ if $x<0$ and $\ln$ is only defined for positive values. 
Since the question have tag "real-analysis", and the expression $(x+\sin x)^{\tan x}$ is complex for (most) negative $x$, I therefore suggest that one should calculate the one-sided limit
$$
\lim_{t\to 0^+}(x+\sin x)^{\tan x}
$$
only.
There are already many correct answers doing this (if one assumes $x>0$ in the solutions), so I won't write yet another one, but let me just point out that before going into too complicated calculations, we can use intuition:
When $x$ is small $\sin x\approx x$ and $\tan x\approx x$, so one would expect the limit to be the same as the one of $(x+x)^x$. But 
$$
(2x)^x=e^{x\ln(2x)}.
$$
Since $x\ln(2x)\to 0$ as $x\to0^+$ and the exponential function is continuous at $0$ (with value 1), the limit should have value $1$. I emphasize that this is not a proof, but it could be made into a proof by expanding $\sin x$ and $\tan x$ and keeping the error term.
A: Just as r9m commented, take logarithms of both sides. So, starting with $$A=\big(x+\sin (x)\big)^{\tan (x)}$$ $$\log(A)=\tan(x) \log\big(x+\sin (x)\big)=\tan(x) \Big(\log(x)+\log\big(1+\frac{\sin (x)}{x}\big)\Big)$$ Now, remember that, close to $x=0$, $\tan(x)=x+O\left(x^2\right)$.  The second logarithm tends to $\log(2)$; so $$\log(A)\approx x\big(\log(x)+\log(2)\big)=x\log(x)+x\log(2)$$ I am sure that you can finish from here.
